Transformation of uniform random variables: Stuck on a question about density functio

**celestial** · December 12th 2011, 08:30 AM

Given: X and Y, independent, continuous uniform random variables on the interval (0,1)

Goal: Find the density function of $\frac{X}{X+Y}$

My approach:
Use the transformation technique (as in point 5 of this wikipedia page Integration by substitution - Wikipedia, the free encyclopedia).

Define $Z_{1} = \frac{X}{X+Y}$ . Define $Z_{2} = X$ .
I find that the joint density of $Z_{1}$ and $Z_{2}$ is

$f_{Z_{1},Z_{2}}(z_{1},z_{2}) = \frac{Z_{2}}{{Z_{1}}^{2}}$

And this is where i get stuck... I tried integrating $Z_{2}$ out of it but nothing reasonable comes out (not a density function anyway). So two questions:

Did I find the joint density correct?
I tried integrating $Z_{2}$ over the interval $Z_{1}$ to 1, is this correct?
What should be next step?

Thanks for any help!

**chisigma** · December 12th 2011, 10:28 AM

Originally Posted by celestial

Given: X and Y, independent, continuous uniform random variables on the interval (0,1)

Goal: Find the density function of $\frac{X}{X+Y}$

My approach:
Use the transformation technique (as in point 5 of this wikipedia page Integration by substitution - Wikipedia, the free encyclopedia).

Define $Z_{1} = \frac{X}{X+Y}$ . Define $Z_{2} = X$ .
I find that the joint density of $Z_{1}$ and $Z_{2}$ is

$f_{Z_{1},Z_{2}}(z_{1},z_{2}) = \frac{Z_{2}}{{Z_{1}}^{2}}$

And this is where i get stuck... I tried integrating $Z_{2}$ out of it but nothing reasonable comes out (not a density function anyway). So two questions:

Did I find the joint density correct?
I tried integrating $Z_{2}$ over the interval $Z_{1}$ to 1, is this correct?
What should be next step?

Thanks for any help!

May be that it is useful to devide the procedure into two steps. The first step is to find the p.d.f. of the random variable $Z=\frac{X}{Y}$ being X and Y independent random variables uniformely distributed in $[0,1]$ . Proceeding as in...

http://www.mathhelpforum.com/math-he...es-193906.html

... we find for $0<z<1$ ...

$p_{z}(z)= \int_{0}^{1} y\ dy =\frac{1}{2}$ (1)

... for $z>1$ ...

$p_{z}(z)= \int_{0}^{\frac{1}{z}} y\ dy =\frac{1}{2\ z^{2}}$ (2)

... and for $z<0$ ...

$p_{z}(z)= 0$ (3)

The second step is to find the p.d.f. of the random variable $U=\frac{1}{1+Z}$ ... how to proceed?...

Marry Christmas from Serbia

$\chi$ $\sigma$

**celestial** · December 12th 2011, 11:44 AM

Originally Posted by chisigma

May be that it is useful to devide the procedure into two steps. The first step is to find the p.d.f. of the random variable $Z=\frac{X}{Y}$ being X and Y independent random variables uniformely distributed in $[0,1]$ . Proceeding as in...

http://www.mathhelpforum.com/math-he...es-193906.html

... we find for $0<z<1$ ...

$p_{z}(z)= \int_{0}^{1} y\ dy =\frac{1}{2}$ (1)

... for $z>1$ ...

$p_{z}(z)= \int_{0}^{\frac{1}{z}} y\ dy =\frac{1}{2\ z^{2}}$ (2)

... and for $z<0$ ...

$p_{z}(z)= 0$ (3)

The second step is to find the p.d.f. of the random variable $U=\frac{1}{1+Z}$ ... how to proceed?...

Marry Christmas from Serbia

$\chi$ $\sigma$

Nice trick with the U! Thanks for that. I knew the ratio distribution of two standard uniforms but couldn't get any use if it. So let me see if Mary and Jesus have answered my prayers and blessed me with knowledge:

So using the ratio of two uniform variables, let's call $g(z) = M = \frac{1}{1+Z}$ . Assume we can prove all the desirable properties like g being a bijective function, and its inverse having a continuous derivative (I haven't proven them yet), then I have $g^{-1}(M) = \frac{1-M}{M}$ . Hence I have for $f_{M}(m) = \frac{ \partial g^{-1}(m)}{\partial m}f_{z}(g^{-1}(m))$
So then I'll have

for $0 < \frac{1-m}{m} < 1$

$f_{M}(m) = - \frac{1}{2 \cdot m^{2}}$

for $\frac{1-m}{m} > 1$

$f_{M}(m) = - \frac{1}{2 \cdot m^{2} \cdot \left(\frac{1-m}{m}\right)}$

and 0 otherwise?

Of course I need to simplify the conditions on m (the intervals) , but is that the right direction?

**chisigma** · December 13th 2011, 12:07 AM

A preminary investigation on the functional equation...

$U=\frac{1}{1+Z},\ Z \ge 0$ (1)

... may be useful. From (1) it is clear that for $u<0$ is $P\{U<u\}=0$ and for $u>0$ is $P\{U<u\}=1$ , so that we have to investigate what happens for $0<u<1$ . From (1) we derive...

$P\{U<u\} = P\{Z>\frac{1-U}{U}\}= \int_{\frac{1-u}{u}}^{\infty} p_{z}(z)\ dz$ (2)

The function $p_{z}(*)$ has been computed in a previous post...

$p_{z}(z)=\begin{cases}\frac{1}{2} &\text{if}\ 0<z<1\\ \frac{1}{2\ z^{2}} &\text{if}\ z>1\\ 0 &\text{elsewhere}\end{cases}$ (3)

Now deriving (2) and taking into account (3) we obtain...

$p_{u}(u)=\begin{cases}\frac{1}{2\ (1-u)} &\text{if}\ 0<z<\frac{1}{2}\\ \frac{1}{2\ u^{2}} &\text{if}\ \frac{1}{2}<z<1\\ 0 &\text{elsewhere}\end{cases}$ (4)

Marry Christmas from Serbia

$\chi$ $\sigma$

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